A 5.00 g bullet is fired with a velocity of118.5 m/s toward a 10.80 kg stationar

Question

A 5.00 g bullet is fired with a velocity of118.5 m/s toward a 10.80 kg stationary solid block resting on africtionless surface.
(a) What is the change in momentum of the bulletif it is embedded in the block?
(b) What is the change in momentum of the bullet if it ricochets inthe opposite direction with a speed of 99 m/s?

A 5.00 g bullet is fired with a velocity of118.5 m/s toward a 10.80 kg stationary solid block resting on africtionless surface.
(a) What is the change in momentum of the bulletif it is embedded in the block?
(b) What is the change in momentum of the bullet if it ricochets inthe opposite direction with a speed of 99 m/s?

Explanation / Answer

m = 5 g, M = 10.80 kg, v = 118.5 m/s According to conservation of energy    mv = (M + m)V        V = mv/(M + m)           = 0.5925 /10.805           = 0.05483m/s Therefore change in momentum of the bullet = m(v - V)          = 0.005 * (118.5 -0.05438)                                                                    = 0.5922 kgm/s If v' = -99 m/s Then change in momentum of the bullet = m(v - v')                                                             = 0.005 * (118.5 + 99)             = 1.0875 kgm/s

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